class Solution {
public:
int waysToSplitArray(vector<int>& nums) {
// Keep track of sum of elements on left and right sides
long long leftSum = 0, rightSum = 0;
// Initially all elements are on right side
for (int num : nums) {
rightSum += num;
}
int count = 0;
// Try each possible split position
for (int i = 0; i < nums.size() - 1; i++) {
// Move current element from right to left side
leftSum += nums[i];
rightSum -= nums[i];
// Check if this creates a valid split
if (leftSum >= rightSum) {
count++;
}
}
return count;
}
};
You are given a 0-indexed integer array nums
of length n
.
nums
contains a valid split at index i
if the following are true:
- The sum of the first
i + 1
elements is greater than or equal to the sum of the lastn - i - 1
elements. - There is at least one element to the right of
i
. That is,0 <= i < n - 1
.
Return the number of valid splits in nums
.
Example 1:
Input: nums = [10,4,-8,7] Output: 2 Explanation: There are three ways of splitting nums into two non-empty parts: - Split nums at index 0. Then, the first part is [10], and its sum is 10. The second part is [4,-8,7], and its sum is 3. Since 10 >= 3, i = 0 is a valid split. - Split nums at index 1. Then, the first part is [10,4], and its sum is 14. The second part is [-8,7], and its sum is -1. Since 14 >= -1, i = 1 is a valid split. - Split nums at index 2. Then, the first part is [10,4,-8], and its sum is 6. The second part is [7], and its sum is 7. Since 6 < 7, i = 2 is not a valid split. Thus, the number of valid splits in nums is 2.
Example 2:
Input: nums = [2,3,1,0] Output: 2 Explanation: There are two valid splits in nums: - Split nums at index 1. Then, the first part is [2,3], and its sum is 5. The second part is [1,0], and its sum is 1. Since 5 >= 1, i = 1 is a valid split. - Split nums at index 2. Then, the first part is [2,3,1], and its sum is 6. The second part is [0], and its sum is 0. Since 6 >= 0, i = 2 is a valid split.
Constraints:
2 <= nums.length <= 10<sup>5</sup>
-10<sup>5</sup> <= nums[i] <= 10<sup>5</sup>